Physics 2022 JAMB Past Questions - Question 9: A long jumper leaves the ground at an angle of 20 degrees above the horizontal and at a speed of 11m/s. How far does it jump in the horizontal direction?

Correct Answer

B

Explanation

To find out how far the long jumper jumps in the horizontal direction, we can use the horizontal component of the initial velocity and the time of flight. The horizontal component of the initial velocity can be found using the formula:

\(v_{i_x} = v_i \times \cos(\theta)\)

Where:
\(v_{i_x}\) = horizontal component of initial velocity
\(v_i\) = initial velocity (11 m/s)
\(\theta\) = angle of takeoff (20 degrees)

\(v_{i_x} = 11 \times \cos(20^\circ)\)
\(v_{i_x} \approx 10.36 \, \text{m/s}\)

The time of flight can be found using the formula:

\(t = \frac{2 \times v_i \times \sin(\theta)}{g}\)

Where:
\(t\) = time of flight
\(v_i\) = initial velocity (11 m/s)
\(\theta\) = angle of takeoff (20 degrees)
\(g\) = acceleration due to gravity (9.81 m/s\(^2\))

\(t = \frac{2 \times 11 \times \sin(20^\circ)}{9.81}\)
\(t \approx 1.85 \, \text{s}\)

Now, we can find the horizontal distance using the formula:

\(d = v_{i_x} \times t\)

\(d = 10.36 \times 1.85\)
\(d \approx 19.18 \, \text{m}\)

So, the long jumper jumps approximately 19.18 meters in the horizontal direction.